Question

Two coils A and B are wound on the same iron core as shown in the figure.the number of turns in A and B are Na and Nb respectively. What is the ratio of magnetic flux linked with them. Please do not spam. Please.

Answer 1

Given:

Two coils A and B are wound on the same iron core. The number of turns in A and B are Na and Nb respectively.

To find:

Ratio of magnetic flux linked with the coil.

Calculation:

First coil :

$\therefore \: \phi = B \times A$

$= > \: \phi = \dfrac{ N_{A}\mu_{0} (I_{A})}{2\pi r} \times A$

$= > \: \phi = \dfrac{ N_{A}\mu_{0} (I_{A})A}{2\pi r}$

For the 2nd coil :

$\therefore \: \phi = B \times A$

$= > \: \phi = \dfrac{ N_{B}\mu_{0} (I_{B})}{2\pi r} \times A$

$= > \: \phi = \dfrac{ N_{B}\mu_{0} (I_{B})A}{2\pi r}$

So, required ratio :

$\boxed{ \sf{ \phi1 : \phi2 = N_{A}I_{A} : N_{B}I_{B}}}$